g ( x ) = ∫ 3 x e t 2 − t d t {\displaystyle g(x)=\int _{3}^{x}e^{t^{2}-t}dt} d d x g ( x ) = d d x [ ∫ 3 x e t 2 − t d t ] = 1 e x 2 − x − 0 e 3 2 − 3 = e x 2 − x {\displaystyle {\frac {d}{dx}}g(x)={\frac {d}{dx}}\left[\int _{3}^{x}e^{t^{2}-t}dt\right]=1e^{x^{2}-x}-0e^{3^{2}-3}=e^{x^{2}-x}} Therefore, g ′ ( x ) = e x 2 − x {\displaystyle g'(x)=e^{x^{2}-x}}
![{\displaystyle {\frac {d}{dx}}g(x)={\frac {d}{dx}}\left[\int _{3}^{x}e^{t^{2}-t}dt\right]=1e^{x^{2}-x}-0e^{3^{2}-3}=e^{x^{2}-x}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/ef9035dbef08f314dec1abe563b675abf3af17c2)